Fundamentals of Matrix Algebra, 2011a

Chapter 4
Eigenvalues and Eigenvectors
. Example 89 .Find the eigenvalues of A, and for each eigenvalue, give one eigenvector,
where
⎡
A = ⎣ 2 −1 1 ⎤
0 1 6 ⎦ .
0 3 4
S We first compute the characterisc polynomial, set it equal to 0,
then solve for λ. We’ll use cofactor expansion down the first column (since it has lots
of zeros).
det (A − λI) =
∣
2 − λ −1 1
0 1 − λ 6
0 3 4 − λ
∣ 1 − λ 6
3 4 − λ
= (2 − λ)
= (2 − λ)(λ 2 − 5λ − 14)
= (2 − λ)(λ − 7)(λ + 2)
∣
∣
Noce that while the characterisc polynomial is cubic, we never actually saw a
cubic; we never distributed the (2 − λ) across the quadrac. Instead, we realized that
this was a factor of the cubic, and just factored the remaining quadrac. (This makes
this example quite a bit simpler than the previous example.)
Our eigenvalues are λ 1 = −2, λ 2 = 2 and λ 3 = 7. We now find corresponding
eigenvectors.
For λ 1 = −2:
We need to solve the equaon (A − (−2)I)⃗x = ⃗0. To do this, we form the appropriate
augmented matrix and put it into reduced row echelon form.
⎡
⎣ 4 −1 1 0 ⎤ ⎡
0 3 6 0 ⎦
−→ rref ⎣ 1 0 3/4 0 ⎤
0 1 2 0 ⎦
0 3 6 0
0 0 0 0
Our soluon, in vector form, is
⎡
⃗x = x 3
⎣ −3/4 ⎤
−2 ⎦ .
1
We can pick any nonzero value for ⎡ x 3 ; a nice choice would get rid of the fracons.
So we’ll set x 3 = 4 and choose ⃗x 1 = ⎣ −3 ⎤
−8 ⎦ as our eigenvector.
4
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